제출 #242562

#제출 시각아이디문제언어결과실행 시간메모리
242562VimmerBurza (COCI16_burza)C++14
0 / 160
1098 ms65656 KiB
#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("fast-math")
//#pragma GCC optimize("no-stack-protector")

#define F first
#define S second
#define sz(x) int(x.size())
#define pb push_back
#define N 405
#define M ll(1e9 + 7)
#define inf 1e9 + 1e9

using namespace std;
//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef short int si;
typedef array <int, 2> a2;

//typedef tree <int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

vector <int> g[N];

int n, k, h[N], mx_h[N];

bool f[N][N][N], dp[N][N];

void dfs(int v, int p)
{
    if (p != -1) h[v] = h[p] + 1;

    mx_h[v] = h[v];

    if (h[v] == k)
        {
            for (int i = 0; i < N; i++)
              for (int j = 0; j < N; j++) f[v][i][j] = 0;

            for (int i = 0; i < k; i++) f[v][i][1] = 1;

            return;
        }

    for (auto it : g[v])
    {
        if (it == p) continue;

        dfs(it, v);

        mx_h[v] = max(mx_h[v], mx_h[it]);

        if (mx_h[it] >= k)
        {
            memset(dp, 0, sizeof(dp));

            for (int i = 0; i < k; i++)
                for (int j = 0; j <= k; j++)
                  if (f[v][i][j])
                  {
                      for (int u = i + j; u < k; u++)
                        for (int kl = 0; kl + j <= k; kl++)
                          if (f[it][u][kl]) dp[i][kl + j] = 1;
                  }

            for (int i = 0; i < k; i++)
                for (int j = 0; j <= k; j++) f[v][i][j] = dp[i][j];
        }
    }

    for (int i = 0; i < h[v]; i++) f[v][i][1] = 1;
}

int main()
{
    //freopen("input.txt", "r", stdin); //freopen("output4.txt", "w", stdout);

    ios_base::sync_with_stdio(0); istream::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n >> k;

    for (int i = 1; i < n; i++)
    {
        int x, y;

        cin >> x >> y;

        g[x].pb(y);

        g[y].pb(x);
    }

    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
          for (int u = 0; u < N; u++) f[i][j][u] = 1;

    dfs(1, -1);

    for (int j = 0; j <= k; j++)
      if (f[1][0][j]) {cout << "DA" << endl; exit(0);}

    cout << "NE" << endl;
}
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